Answer:

Given:
- Length of the steel cylindrical shaft (L) = 3m
- Location of pulley from one end (x) = 1m

To find:
Ratio of maximum shearing stresses developed in 1m and 2m lengths of the shaft.

Assumptions:
- The shaft is made up of homogeneous material.
- The shaft is subjected to pure torsion.
- The shaft is a solid circular shaft.

Formulae:
- Polar moment of inertia (J) = π/32 * d^4
- Shear stress (τ) = T*r/J
- Torsional rigidity (GJ) = Tθ/γ
- Maximum shear stress (τmax) = T*rmax/J

Where,
d = diameter of the shaft
r = radius of the shaft
T = applied torque
θ = angle of twist
γ = angle of twist per unit length
rmax = maximum radius
J = polar moment of inertia

Calculation:
Let us consider 1m and 2m sections of the shaft separately.

1m section:
- Length of the shaft (L) = 1m
- Location of pulley from one end (x) = 1m
- Length of the section (L1) = 1m
- Length of the remaining section (L2) = 2m

- Polar moment of inertia (J) = π/32 * d^4
- Shear stress (τ) = T*r/J
- Torsional rigidity (GJ) = Tθ/γ

Now, let us calculate the maximum shear stress developed in the 1m section.

- Maximum shear stress (τmax) = T*rmax/J

For a solid circular shaft, rmax = r.

- Maximum shear stress (τmax) = T*r/J

Since the shaft is rigidly held at both ends, the angle of twist (θ) is given by,

- θ = TL/GJ

Substituting the values of L1, d and θ in the above equation, we get,

- θ1 = TL1/GJ

Hence, the maximum shear stress developed in the 1m section is given by,

- τmax1 = T*r/J = T*0.5/d^3
- θ1 = TL1/GJ = T*1/GJ

2m section:
- Length of the section (L2) = 2m

- Polar moment of inertia (J) = π/32 * d^4
- Shear stress (τ) = T*r/J
- Torsional rigidity (GJ) = Tθ/γ

Now, let us calculate the maximum shear stress developed in the 2m section.

- Maximum shear stress (τmax) = T*rmax/J

For a solid circular shaft, rmax = r.

- Maximum shear stress (τmax) = T*r/J

Since the shaft is rigidly held at both ends, the angle of twist (θ) is given by,

- θ = TL/GJ

Substituting the values of L2, d and θ in the above equation, we get,

- θ